what is the antiderivative 1/(1 + 4x^2)?

Answer 1

user this:the antiderivative 1/(a^2+y^2)=(1/a)arctan(y/a)
1/(1+4x^2)=(1/4)/(1/4+x^2) ;then a=1/2 ,y=x
the antiderivative 1/(1+4x^2)=[(1/4)/(1/2)]arctan(2x)=(1/2)arctan(2x)

Answer 2

Consider this,
y= arctan x
x = tan y
dx/dy = sec^2 y = 1 + tan^2 y = 1+x^2
then,
dy/dx = 1/(1+x^2)

Now work the antiderivative of the required function

Answer 3

1/(1+4x^2)=(1+4x^2)^-1

antiderivative=1/2arctg(2x)+C